Question: Simplify the following expression and state the condition under which the simplification is valid. You can assume that $r \neq 0$. $n = \dfrac{7(r - 9)}{-3} \times \dfrac{-4}{r^2 - 9r} $
Answer: When multiplying fractions, we multiply the numerators and the denominators. $n = \dfrac{ 7(r - 9) \times -4 } { -3 \times (r^2 - 9r) } $ $ n = \dfrac {-4 \times 7(r - 9)} {-3 \times r(r - 9)} $ $ n = \dfrac{-28(r - 9)}{-3r(r - 9)} $ We can cancel the $r - 9$ so long as $r - 9 \neq 0$ Therefore $r \neq 9$ $n = \dfrac{-28 \cancel{(r - 9})}{-3r \cancel{(r - 9)}} = -\dfrac{28}{-3r} = \dfrac{28}{3r} $